A peice of ice weighting 5g at -20 centigrade is put into 10g of ice at 30C. Assuming no heat is lost to the outside.Calculate the final tempature of the mixture?

Answer 1

Ice at 30 deg C??? Not likely but anyway

specific heat of ice (about 2 J/gram deg c) does not matter in the end here
call it sh
heat into cold ice = heat out of hot ice
5 (T - -20) * sh = 10 (30-T) * sh
5 T + 100 = 300 - 10 T
15 T = 300
T = 20
BUT I doubt the whole thing
suspect that 10 grams was water at 30 deg
In that case
Heat in to heat the ice up to zero
5 (20)*sh
Heat in to melt the ice
5 * hm where hm is the heat of fusion of water
Heat in to warm melted ice up to T
5(T-0) * shw
where shw is specific heat of water

Heat out of 30 deg water
10(30-T)*shw
so
5 (20)*sh + 5 * hm + 5(T-0) * shw = 10(30-T)*shw
If T comes out below zero, then T = 0 and some of the 30 deg water froze